Simple Lie algebras and ideals

  • #26
matt grime
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Don Aman said:
It seems you consider the universal enveloping algebra "contrived", but as it is a universal functor from the category of Lie algebras, it is natural in the usual sense of the word.
it is universal in the usual sense of the word. but embedding a (small) lie algebra into an (infinite dimensional) quotient of a tensor algebra does not endow the original lie algebra with a 'naturally occuring' product, does it? the product so given is not (necessarily) defined on the lie algebra before you make this construction. the example i gave of C^3 with relations forcing it to be sl_2 shows that. C^3 comes with no 'natural' algebra structure for which [ef]=ef-fe.

we can embed (finitely generated) abelian monoids into (finitely generated abelian) groups (at least thati s my gut reaction, i can't say i've proved it, but it seems clear), that doesn't mean that there isn't a naturally occuring monoid that isn't a group, does it?
But every Lie algebra has a multiplication, and the only way you'll find one without multiplication is if you turn a blind eye to it.
and the question was if there was a naturally occuring situation when we *do* turn a blind eye to it, or more accurately when we plain don't see it to begin with (like N in Z). why is that so hard to grasp? i can't think of one, and you can't either, apparently, so why do you keep telling me you can't think of one?
That's why I mention the Lie algebras of Lie groups that aren't matrix groups (say, Spin(n) for a big enough n). The only way it'll have a multiplication is if you go out of your way to define one.
so now you can think of one? which is it?

As another example I know fundamental group of a space can be made arbitrary, that doesn't mean every group is a fundamental group, it is merely isomorphic to one. Is the distinction clear?
 
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  • #27
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matt grime said:
As another example I know fundamental group of a space can be made arbitrary, that doesn't mean every group is a fundamental group, it is merely isomorphic to one. Is the distinction clear?
Of course the distinction is clear. Every (fd) Lie algebra is isomorphic to a matrix algebra, which has a natural product. But that doesn't mean it is a matrix algebra, in exactly the same sense that not every group is a fundamental group. I mention now, for the third time, an example of a Lie algebra that is not a matrix algebra, and therefore does not have your product: the Lie algebra of a Lie group which is not a matrix group. It is of course isomorphic to a matrix algebra, and so in that sense has a product, but then again, so is every (fd) Lie algebra. It is of course embeddable in its universal enveloping algebra, and so in that sense has product, but then again, so is every Lie algebra.

Are there Lie algebras without products? Sure, simply don't include a product. Are there Lie algebras which are not isomorphic to algebras with product? No. As far as I can tell, the answer to your question is either "obviously yes" or "obviously no", depending on your view.
 
  • #28
matt grime
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Don Aman said:
Are there Lie algebras without products? Sure, simply don't include a product. Are there Lie algebras which are not isomorphic to algebras with product? No. As far as I can tell, the answer to your question is either "obviously yes" or "obviously no", depending on your view.


which is why i included the words 'naturally occuring', that was sort of the key point in the question that didn't make it vacuous. indeed i gave a formal example of a lie algebra that didn't come with a product

however, as for the example given two tangents at the identity what is [ab] of them? seriously, i'm an algebraist, know nothing of differential geometry, i thought the tangent space was the span of the operators [tex]\partial_x[/tex] which do have a natural product since they are in the associative algebra of differential operators, and that the bracket was ab-ba in this space.

now, if as is entirely possible, there is another definition of tangent space that doesn't require you to evaluate things in this algebra, but which takes in a,b and spits out [ab] naturally then that'd be what the question wanted to know about.
 
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  • #29
George Jones
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In brief: the Lie algebra of a Lie Group is the Lie algebra of left-invariant vector fields, and, since a left-invariant vector field is uniquely determined by its value in the tangent space of the identity, the commutator product of left-invariant vector can be used to give the tangent space of the identity Lie algbera structure.

Now the details.

Let [itex]M[/itex] be an n-diemnsional differentiable manifold, and let [itex]f: M \rightarrow M[/itex] be a diffeomorphism. [itex]f[/itex] can be used to define for any [itex]p \in M[/itex] a pushforward map (differential) [itex]f_{*} : T_{p} \rightarrow T_{f\left(p\right)}[/itex] between tangent spaces as follows.

Suppose [itex]p \in M[/itex] is in the image of a curve [itex]\gamma: I \rightarrow M[/itex], where [itex]I[/itex] is an interval in [itex]\mathbb{R}[/itex]. Then, [itex]\gamma' = f \circ \gamma[/itex] is a curve that has [itex]f\left(p\right)[/itex] in its image. Let [itex]v[/itex] be the tangent vector to [itex]\gamma[/itex] at [itex]p[/itex], and define [itex]f_*[/itex] by taking [itex]f_{*} \left( v \right)[/itex] to be the tangent vector to [itex]\gamma'[/itex] at [itex]f \left( p \right)[/itex].

This is the natural definition of [itex]f_*[/itex] to use with the equivalence classes of curves definition of tangent vectors, and the equivalent definition of [itex]f_*[/itex] for the derivation definition of tangent vectors shows that every [itex]h_*[/itex] is linear.

Now let [itex]M[/itex] also have Lie group structure. For any [itex]g \in M[/itex], define a diffeomorphism (action) [itex]l_{g} : M \rightarrow M[/itex] by [itex]l_{g} h = gh[/itex] for every [itex]h \in M[/itex].

Let [itex]X[/itex] be a vector field on [itex]M[/itex], and [itex]X_h[/itex] be its representative in the tangent space [itex]T_h[/itex] at [itex]h \in M[/itex]. Use [itex]l_{g*}[/itex] to push forward to a vector [itex]l_{g*} X_{h}[/itex] in [itex]T_{l_g h} = T_{gh}[/itex]. But [itex]X[/tex] itself has a representative[itex]X_{l_g h}[/itex] in [itex]T_{l_g h} = T_{gh}[/itex].

The vector field [itex]X[/itex] is said to be left-invariant if [itex]l_{g*} X_{h} = X_{l_{g} h}[/itex] for every [itex]h \in M[/itex]. It is easy to show that any left-invariant vector field is determined by its representative at the identity. Also, if [itex]X[/itex] and [itex]Y[/itex] are 2 left-invariant vector fields, then so is the vector field [itex]\left[ X , Y \right][/itex].

Let [itex]X_e[/itex] and [itex]Y_e[/itex] be arbitrary tangent vectors in the tangent space of the identity, and let [itex]X[/itex] and [itex]Y[/itex] be the unigue vector fields that they determine. Define [itex]\left[ X_{e}, Y_{e} \right][/itex] by

[tex]\left[ X_{e}, Y_{e} \right] = \left[ X , Y \right]_{e}.[/tex]

This turns the tangent space at the indentity into a Lie algebra isomorphic to the Lie algebra of left-invariant vector fields. Members of the Lie algebra of left-invariant vector fields have natural compositions as vector fields (that aren't vector fields), but I don't see how to extend this to an associative product on the tangent space at the identity.

So, it seems that for a general Lie group, the Lie algebra of left-invariant vector fields doesn't meet your criteria, but the Lie algebra of the tangent space at the identity might meet your criteria. ???

Regards,
George
 
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  • #30
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matt grime said:
however, as for the example given two tangents at the identity what is [ab] of them?
tangent vectors are first order differential operators. the product of two of those is a second order differential operator.
 

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